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ch01

Differential Equations: An Introduction to Modern Methods and Applications

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Chapter 1 Section 1.1 1. –4 –2 0 2 4 y(t) –4 –2 2 4 t For y > 3 / 2, the slopes are negative, and, therefore the solutions decrease. For y < 3 / 2, the slopes are positive, and, therefore, the solutions increase. As a result, y 3 / 2 as t → ∞ 2. –4 –2 0 2 4 y(t) –4 –2 2 4 t For y > 3 / 2, the slopes are positive, and, therefore the solutions increase. For y < 3 / 2, the slopes are negative, and, therefore, the solutions decrease. As a result, y diverges from 3 / 2 as t → ∞ 3. 1

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–4 –2 0 2 4 y(t) –4 –2 2 4 t For y > - 3 / 2, the slopes are positive, and, therefore the solutions increase. For y < - 3 / 2, the slopes are negative, and, therefore, the solutions decrease. As a result, y diverges from - 3 / 2 as t → ∞ 4. –4 –2 0 2 4 y(t) –4 –2 2 4 t For y > - 1 / 2, the slopes are negative, and, therefore the solutions decrease. For y < - 1 / 2, the slopes are positive, and, therefore, the solutions increase. As a result, y → - 1 / 2 as t → ∞ 5. –4 –2 0 2 4 y(t) –4 –2 2 4 t 2
For y > - 1 / 2, the slopes are positive, and, therefore the solutions increase. For y < - 1 / 2, the slopes are negative, and, therefore, the solutions decrease. As a result, y diverges from - 1 / 2 as t → ∞ 6. –4 –2 0 2 4 y(t) –4 –2 2 4 t For y > - 2, the slopes are positive, and, therefore the solutions increase. For y < - 2, the slopes are negative, and, therefore, the solutions decrease. As a result, y diverges from - 2 as t → ∞ 7. For the solutions to satisfy y 3 as t → ∞ , we need y 0 < 0 for y > 3 and y 0 > 0 for y < 3. The equation y 0 = 3 - y satisfies these conditions. 8. For the solutions to satisfy y 2 / 3 as t → ∞ , we need y 0 < 0 for y > 2 / 3 and y 0 > 0 for y < 2 / 3. The equation y 0 = 2 - 3 y satisfies these conditions. 9. For the solutions to satisfy y diverges from 2, we need y 0 > 0 for y > 2 and y 0 < 0 for y < 2. The equation y 0 = y - 2 satisfies these conditions. 10. For the solutions to satisfy y diverges from 1 / 3, we need y 0 > 0 for y > 1 / 3 and y 0 < 0 for y < 1 / 3. The equation y 0 = 3 y - 1 satisfies these conditions. 11. –4 –2 0 2 4 y(t) –4 –2 2 4 t 3

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y = 0 and y = 4 are equilibrium solutions; y 4 if initial value is positive; y diverges from 0 if initial value is negative. 12. –4 –2 0 2 4 6 y(t) –4 –2 2 4 t y = 0 and y = 5 are equilibrium solutions; y diverges from 5 if initial value is greater than 5; y 0 if initial value is less than 5. 13. –4 –2 0 2 4 y(t) –4 –2 2 4 t y = 0 is equilibrium solution; y 0 if initial value is negative; y diverges from 0 if initial value is positive. 14. 4
–4 –2 0 2 4 y(t) –4 –2 2 4 t y = 0 and y = 2 are equilibrium solutions; y diverges from 0 if initial value is negative; y 2 if initial value is between 0 and 2; y diverges from 2 if initial value is greater than 2. 15. (j) 16. (c) 17. (g) 18. (b) 19. (h) 20. (e) 21. (a) Let q ( t ) denote the amount of chemical in the pond at time t . The chemical q will be measured in grams and the time t will be measured in hours. The rate at which the chemical is entering the pond is given by 300 gallons/hour · . 01 grams/gal = 300 · 10 - 2 . The rate at which the chemical leaves the pond is given by 300 gallons/hour · q/ 1 , 000 , 000 grams/gal = 300 · q 10 - 6 . Therefore, the differential equation is given by

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ch01 - Chapter 1 Section 1.1 1 4 y(t 2 0 2 4 4 2 2 t 4 For...

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